Remove Nth Node From End of List "移除链表中的倒数第N项"

Given a linked list, remove the nth node from the end of list and return its head.

For example,

Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will
always be valid.

Try to do this in one pass.

移除链表的倒数第N项

样例：

[1,2]

1[/b]

[1]

[]

思路：此题比较简单，采用窗口思想，窗大小比n大1，这样方便删除节点操作。滑动窗口，当窗的右侧到末尾时，左侧的下一个节点删除即可
















/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
ListNode nullHead = new ListNode(0);
nullHead.next = head;//为方便操作，自己添加一个头
ListNode l = nullHead;
ListNode r = nullHead;
try{
int i = n;
while(i>0){
r = r.next;
i--;
}//设置窗口
}catch(Exception e){
return null;//n不合理时（过大）
}
while(r.next!=null){
r = r.next;//窗口滑动至队尾
l = l.next;
}
try{
l.next = l.next.next;
}catch(Exception e){//n不合理时（过大）
return null;
}

return nullHead.next;//去掉自己添加的头
}
}